Question: Simplify the following expression and state the condition under which the simplification is valid. $r = \dfrac{-3y^2 + 243}{y^3 + 19y^2 + 90y}$
First factor out the greatest common factors in the numerator and in the denominator. $ r = \dfrac {-3(y^2 - 81)} {y(y^2 + 19y + 90)} $ $ r = -\dfrac{3}{y} \cdot \dfrac{y^2 - 81}{y^2 + 19y + 90} $ Next factor the numerator and denominator. $ r = - \dfrac{3}{y} \cdot \dfrac{(y + 9)(y - 9)}{(y + 9)(y + 10)}$ Assuming $y \neq -9$ , we can cancel the $y + 9$ $ r = - \dfrac{3}{y} \cdot \dfrac{y - 9}{y + 10}$ Therefore: $ r = \dfrac{ -3(y - 9)}{ y(y + 10)}$, $y \neq -9$